Prove combinatorically, that for $n \geq m$: $\sum_{k=0}^m \dbinom{n}{k} \dbinom{n-k}{m-k}= 2^m \dbinom{n}{m}$

Question

Prove combinatorically, that for $n \geq m$: $$\sum_{k=0}^m \dbinom{n}{k} \dbinom{n-k}{m-k}= 2^m \dbinom{n}{m}$$

The right hand side tells that us the number of ways we can pick either white or black $m$ balls from a total of $n$ balls. But I'm not sure what the connection between that and the left hand side is.